1. 已知关于 x,y 的方程组$\{\begin{array}{l} x + 2y = k + 7,\\ 3x + 5y = 4k + 18\end{array} $的解也是方程$2x + 3y = 11$的解,求k的值.
答案：解：$\{\begin{array}{l}x + 2y = k + 7,①\\3x + 5y = 4k + 18,②\end{array} $
②$-$①，得$2x + 3y = 3k + 11$．
$∵$关于$x$，$y$的方程组$\{\begin{array}{l}x + 2y = k + 7\\3x + 5y = 4k + 18\end{array} $的解也是方程$2x + 3y = 11$的解，
$∴3k + 11 = 11$，$∴k = 0$．
2. (2024·盱眙月考)已知关于 x,y 的方程组$\{\begin{array}{l} nx + (n + 1)y = n + 2,\\ x - 2y + mx = - 5\end{array} $(n 是常数).
(1)当$n = 1$时,则方程组可化为$\{\begin{array}{l} x + 2y = 3,\\ x - 2y + mx = - 5.\end{array} $
①请直接写出方程$x + 2y = 3$的所有非负整数解;
②若该方程组的解也满足方程$x + y = 2$,求 m 的值.
(2)当$n = 3$时,如果方程组有整数解,求整数 m 的值.
答案：解：(1)①$∵x$，$y$为非负整数，$∴$方程$x + 2y = 3$的所有非负整数解为$\{\begin{array}{l}x = 1\\y = 1\end{array} $，$\{\begin{array}{l}x = 3\\y = 0\end{array} $．
②根据题意，得$\{\begin{array}{l}x + 2y = 3,①\\x + y = 2,②\end{array} $
①$-$②，得$y = 1$．
将$y = 1$代入②，得$x = 1$，
$∴$原方程组的解是$\{\begin{array}{l}x = 1\\y = 1\end{array} $．
将$\{\begin{array}{l}x = 1\\y = 1\end{array} $代入$x - 2y + mx = - 5$，得$m = - 4$．
(2)当$n = 3$时，原方程组可化为$\{\begin{array}{l}3x + 4y = 5,①\\x - 2y + mx = - 5,②\end{array} $
②$× 2$，得$2x - 4y + 2mx = - 10$，③
①$+$③，得$(5 + 2m)x = - 5$．
$∵$方程组有整数解，且$m$为整数，
$∴5 + 2m = \pm 1$或$5 + 2m = \pm 5$．
当$5 + 2m = 1$时，$m = - 2$，此时方程组的解是$\{\begin{array}{l}x = - 5\\y = 5\end{array} $；
当$5 + 2m = - 1$时，$m = - 3$，此时方程组的解是$\{\begin{array}{l}x = 5\\y = - \dfrac{5}{2}\end{array} $（舍去）；
当$5 + 2m = 5$时，$m = 0$，此时方程组的解是$\{\begin{array}{l}x = - 1\\y = 2\end{array} $；
当$5 + 2m = - 5$时，$m = - 5$，此时方程组的解是$\{\begin{array}{l}x = 1\\y = \dfrac{1}{2}\end{array} $（舍去）．
综上，整数$m$的值为$- 2$或$0$．
